If ^2 = 4 3 , then the general value of is - Vidyadip

MCQ

If ${\sec ^2}\theta = \frac{4}{3}$, then the general value of $\theta $ is

A
$2n\pi \pm \frac{\pi }{6}$
✓
$n\pi \pm \frac{\pi }{6}$
C
$2n\pi \pm \frac{\pi }{3}$
D
$n\pi \pm \frac{\pi }{3}$

✓

Answer

Correct option: B.

$n\pi \pm \frac{\pi }{6}$

b
(b) ${\cos ^2}\theta = \frac{3}{4} = {\cos ^2}\left( {\frac{\pi }{6}} \right) $

$\Rightarrow \theta = n\pi \pm \frac{\pi }{6}$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "MCQ" from Trigonometrical equations→Trigonometrical equations — all question types→MATHS STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

A student was asked to prove a statement P(n) by induction. He proved that P(k + 1) is true whenever P(k) is true for all k > 5 ∈ N and also that P(5) is true. Based on this, he could conclude that P(n) is true:

One root of the equation $\cos x - x + \frac{1}{2} = 0$ lies in the interval

Eccentricity of the ellipse $9{x^2} + 25{y^2} = 225$ is

If the number of five digit numbers with distinct digits and $2$ at the $10^{\text {th }}$ place is $336 \mathrm{k}$, then $\mathrm{k}$ is equal to

Argument of $ - 1 - i\sqrt 3 $ is

Let $\mathrm{ABC}$ be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle $\mathrm{ABC}$ and the same process is repeated infinitely many times. If $\mathrm{P}$ is the sum of perimeters and $Q$ is be the sum of areas of all the triangles formed in this process, then:

$\mathop {\lim }\limits_{\theta \to 0} \frac{{4\theta (\tan \theta - 2\theta \tan \theta )}}{{{{(1 - \cos 2\theta )}^2}}}$ is

Consider all possible permutations of the letters of the word $EARTHQUAKE$ , then the number of permutation containing the word $RAHU$ is

If the sum of the first ten terms of the series ${\left( {1\frac{3}{5}} \right)^2} + {\left( {2\frac{2}{5}} \right)^2} + {\left( {3\frac{1}{5}} \right)^2} + {4^2} + \;\;.\;.\;.\;.\;,$ is $\frac{{16}}{5}m$ then $m$ is equal to :

The sum $1(1!) + 2(2!) + 3(3!) + ....+n (n!)$ equals